Quick quantitative analysis problem

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The discussion revolves around calculating the absolute and relative uncertainties in the final concentration of a NaF solution using two different methods. Method one involves dissolving 0.0042 g of NaF in a 1000 mL volumetric flask, while method two uses 0.0420 g in a 100 mL flask, followed by a 100:1 dilution. Participants noted discrepancies in the calculated concentrations and uncertainties, emphasizing the importance of significant figures and proper unit conversions. Method one was identified as having a higher uncertainty due to the larger volume and smaller mass, while both methods yielded a relative uncertainty of 2%. The conversation highlights the critical nature of accurate measurements and calculations in quantitative analysis.
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Homework Statement


A 1.00 × 10^-4 M NaF solution can be prepared in several ways. Two methods are listed below. The formula weight of NaF is 41.9984 ± 0.0003.

For each method use propagation of error methods to determine the absolute and relative uncertainty in the final concentration.

Method 1
0.0042 ± 0.0001 g of NaF is placed in a 1000.0 ± 0.3 mL volumetric flask. The flask is filled to the mark with water

Method 2
0.0420 ± 0.0001 g of NaF is placed in a 100.00 ± 0.08 mL volumetric flask. The flask is filled to the mark with water. This solution is diluted 100:1 by pipeting 1.000± 0.003 mL of the solution to another 100.00 ± 0.08 mL volumetric flask and then filling to the mark with water.

For each method use propagation of error to determine the absolute and relative uncertainty in the final concentration.

Which method has the most uncertainty? Which individual measurement introduces the most uncertainty?



Homework Equations



Molarity = mol/L

moles = mass/MW

Final Concentration = (Initial Concentration)*(Initial Volume)/(Final Volume)


The Attempt at a Solution


Note: When writing the absolute uncertainties, I purposely write "2." because it needed to be just one significant figure, and writing down just "2" means that there are no significant figures.

For method one, I took the grams of NaF and converted it to moles, which turned out to be 1.0*10^-4 ± 2. * 10^-6 when propagation of uncertainty and the correct significant figures were used. Once I found moles, I then converted to molarity using the following: (1.0*10^-4 ± 2. * 10^-6 )/(1.0*10^3 ± 0.3L) and arrived at 1.0 * 10^-7 ± 2. * 10^-9 M.

Is 1.0 * 10^-7 ± 2. * 10^-9 M the final concentration in this case? I know the problem says it needs to be 1.0 * 10 ^-4 M, but it wanted the percent relative uncertainty, which ended up being 2%. I'm not really sure what else to do regarding method one.



For method two, it seemed a bit easier, but I'm not even sure I did it right. I made use of the following formula: Final Concentration = (Initial Concentration)*(Initial Volume)/(Final Volume).

Since I had found the initial concentration from method one, I just plugged that in and multiplied it by the pipet volume and divided it by the volume of the flask. I arrived at a final concentration, again different from the value that the problem originally stated, of 1.0*10^-9 ± 2. * 10^-11 M with a relative uncertainty of 2% as well. Not sure what else to do...
 
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As a general remark, the value itself should be given with the same "precision" and the same decimal power as the uncertainty. Don't write 1*10-6 ± 1*10-8, write 100*10-8 ± 1*10-8 or 1.00*10-6 ± 0.01*10-6 or (100±1)*10-8.For method 1, you mixed ml and l, so your result is off by a factor of 1000.
2% as relative uncertainty is right.

Since I had found the initial concentration from method one
You do not have that, at least not with the correct uncertainty. The values are different here.
And there is another factor of 100 wrong here.
 

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